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Wed 21 Sep, 2005 06:33 pm
A baseball pitcher throws a fastball at a speed of 44 m/s. The acceleration occurs as the pitcher holds the ball in his hand and moves it through an almost staight-line distance of 3.5 m. Calculate the acceleration, assuming it is uniform. Compare this acceleration to the acceleration due to gravity, 9.80m/s^2.
Please explain how to get to the answer.
Thank you very much for your time.
Classic straight line physics if I read your problem correctly.
The pitcher ramps the balls up to a velocity of 44 m/s in 3.5 m.
The straight line displacement is given by
S=1/2at^2=3.5 m
And final velocity
Vf=at=44 m/s
The common variables are t (time the pitcher accelerated the baseball) and a (acceleration).
Look at the displacement relation
S=1/2at^2=1/2(at)t
So
3.5m=1/2(44 m/s)t & t is the only unknown.
t= (2*3.5 m/(44 m/s)=7/44 s
Now you can substitute t into either relation (displacement or final velocity) to find acceleration
Use final velocity (Vf=at)
44 m/s=at=a*(7/44 m/s)
so
a=Vf/t=44m/s/(7/44 s)=44^2/7 m/s^2=277 m/s^2
Compare this to 1 g (9.8 m/s^2) acceleration
Simply use the ratio of the two accelerations
a/g=277 m/s^2/9.8 m/s^2=28.2 g)
Note all calculations were rounded to 3 significant digits
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